Codeforces Round #456 (Div. 2) Editorial

https://en.wikipedia.org/wiki/Ones%27_complement

17 months ago, # ^ |

+31

Not as fast as my score's decline rate

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elManco

17 months ago, # ^ |

That's sad man :'(

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Flying_Dragon_02

17 months ago, # |

Great editorial! :)

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brdy

17 months ago, # |

Is there a term for binary numbers that fill in all 1's when xor'd? For example 11000 xor 00111 is 11111 so they fulfill this property.

I think if such numbers have a name problem B relies on this idea.

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rop

17 months ago, # ^ |

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o_To_T_oT

17 months ago, # ^ |

I think it's just something like bitwise not. (maybe with specific width)

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brdy

17 months ago, # ^ |

You're right, I made a stupid comment.

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17 months ago, # |

I can't believe B turned out to be so simple, I didn't realise it said "at most k"

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codeforceish

17 months ago, # ^ |

same here but i still cant get solution can u give me its possible examples satisfying the mentioned statement means VALID PROOFING

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http://codeforces.com/contest/912/submission/34650649

17 months ago, # ^ |

Checkout my comment bellow

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codeforceish

17 months ago, # ^ |

can u tell me why the log2(n) is not working here for the calculation of no of bits in the binary rep of n

but when i calculated it recursively through loop and then applying pow(2,no of bits ) It gets accepted

I also use fast mod exponentiation in the case when no of bits gets very large but it doesn't matter i know because for this i set my mod to be very large far beyond then 10^18 so that no overflow occurs

http://codeforces.com/contest/912/submission/34650763 but in the ques pow is still getting accepted due to the less width of the no of bits in the bin rep of n.. Can u plzz tell me why this strange behaviour happening

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wannabe_main

17 months ago, # |

B was simple, now i realize, started this round so late , damn !

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Anadi

17 months ago, # |

← Rev. 2 →

I think you made a small mistake — the complexity in E is O((D(A) + D(B)) * log(D(A) + D(B)) * log10¹⁸)

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17 months ago, # ^ |

I use two pointers so the complexity is true :) We are uploading the sample solutions now.

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Yusif

17 months ago, # |

Can someone explain the calculation of f(x, y) in problem D?

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17 months ago, # ^ |

Just apply all restrictions on both coordinates.

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aadish

17 months ago, # ^ |

Can you explain it little more.

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17 months ago, # ^ |

← Rev. 2 →

Let's solve it for a single x₀. Imagine we're covered by a scoop located at x. What should be applied?

x ≥ 1
x ≤ x0
x + r - 1 ≤ n
x + r - 1 ≥ x₀

Combine them and solve the same problem for y. This will give you a rectangle where each point is valid.

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s3ct4l-r3x

17 months ago, # ^ |

← Rev. 2 →

double fun(int x,int y)

{

double ro,co;

   if(1<= x <= r || n-r+1<=x<= n)ro=min(x,n-x+1);
   else ro= r;

   if(1<= y <= r || m-r+1<=y<= m)co=min( m-y+1 ,y);
   else co= r;

   return ro*co;

}

I came up with this function during the contest which returns the count of scoops in which (x,y) is included. I thought the get(x,y) in editorial gave the same result. But it differs in some cases. Whats wrong with the logic behind my function?

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s3ct4l-r3x

17 months ago, # ^ |

lol! never mind. I dont know what I have been looking at for the past 6 hours. That isnt even a correct implementation. So shameful that i am realizing this after posting it here.

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17 months ago, # ^ |

Shouldn't x0 >= x ?

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17 months ago, # ^ |

Oh, thanks for pointing out.

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noob__coder

17 months ago, # ^ |

← Rev. 2 →

I know that I am asking a silly question but please help me with this. How the given 4 inequality can be combined to get 1. min(n+1,x+r)-max(x,r)

EDIT: I get it now the logic behind this.

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17 months ago, # ^ |

Can you elaborate on the next steps, i.e. how you solve these inequalities for x0 ?

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17 months ago, # |

+12

Very nice problem set! I hope to see more contests from you in future.

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brurrito

17 months ago, # |

For problem D, how do you check if a cell (x,y) has already been visited in O(1) ?

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17 months ago, # ^ |

Make a set of pairs.

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brurrito

17 months ago, # ^ |

Oh okay I'm not really used to use sets that's why ^^ thanks

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yaksha

17 months ago, # ^ |

actually if you go by the solution , it's impossible to get a cell (x , y) two times,.. , but we still need to get the maximum cell at top so that's why we have to use heap or set , whatever you prefer , no worry about them getting repeated

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amandhapola

17 months ago, # |

can somebody explain problem (b) elaborately ?? need help !

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taxi_driver

17 months ago, # ^ |

← Rev. 6 →

Assume that you have number 11 binary representation (1011) , now the maximum xor sum can be 1111 , if you observe more carefully than if you take 1 complement of 11 i.e 0100 and xor it u get 1111 i.e maximum .

So if k==1 you will print n because it is the maximum

else print the maximum 2^p-1 where p are the bits in binary representation of n because 1's complement of n is always less than n

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noob__coder

17 months ago, # ^ |

For k=1 the problem is quite simple. You just print the greatest number which is indeed n. For k>1: [Note that the problem asked us to Select AT MOST k numbes in range n] So we can always select two numbers which will maximize the result. Take the case of n = 14, in binary n = 1110 This contains one zero which can be made 1. How? Take the xor of 1000(in decimal 8) and 0111(in decimal 7) = 1111(in decimal 15) Now 8 is in form = 2^p and 7 is in form = 2^p-1 I hope this helps.

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anubhav_ankit87

4 months ago, # ^ |

Great Explanation.

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17 months ago, # ^ |

Consider the case for n = 4 , k=3, let's write the numbers from 1 to n in binary:

1 => 001
2 => 010
3 => 011
4 => 100

Furthermore, we know that xor of 2 binary digits is is 1 only when both differ ( 1 XOR 0 .. or 0 XOR 1).

After observing the sequence above you realise that 3 XOR 4 is maximum, because all of the digits in both numbers differ

    011
XOR 100
  = 111 (7)

To generalize this we observe that for any n, if k>=2 we can compute the max value by setting all the bits to 1's, since we can guarantee to find 2 numbers whose digits all differ, some examples:

n=11, which is 1011, so answer is 1111, or 15 
n= 33, which is 100001 so the answer is 111111, or 63

to generalise this we can use this formula:

if k ==1:
  answer is n
else:
  answer is 2^(floor(lg(n))+1) - 1

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cyber_ghost02

4 weeks ago, # ^ |

great explanation

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M_H_H_7

17 months ago, # |

the contest was very good for me, but i don't think that was a standard contest.

because normal Div.2 participants mostly solved problems A and B but in a standard contest, a normal participant usually solve 3 problems.

nevertheless thank you for the contest:)

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17 months ago, # |

Auto comment: topic has been updated by rek (previous revision, new revision, compare).

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thedsk

17 months ago, # |

-6

is this contest rated?

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o_To_T_oT

17 months ago, # |

In problem E, how could I estimate |D(A)|? Should I just do what I guest and check it?

( In the contest, I think the size might went to 10^9 or larger, so I didn't try to implement and check it. TAT

BTW, I think it's a good contest, thank you!

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17 months ago, # ^ |

Yep. It's a coding competition after all xD

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aj1

17 months ago, # ^ |

Can you explain how did you come up with "sizes of both D(A) and D(B) do not exceed 106"?

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o_To_T_oT

17 months ago, # ^ |

I think we could just try to implement it and find it out

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apoorvkumar91

17 months ago, # |

Was just curious how long will it take for the ratings to be updated?

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all_out

17 months ago, # |

In problem D, how f(x, y) is formulated ?

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17 months ago, # ^ |

That is "how many scoops cover this point".

For example, (2, 2) from sample case 1 (reference to the pic) has f(2, 2) = 4.

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all_out

17 months ago, # ^ |

I mean, how it is generated?

f(x, y)= (max... — min...)*()

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17 months ago, # ^ |

Take a look at this!

Do such thing with x₀ and y₀ separately and take their cartesian product.

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ricksr

17 months ago, # |

great editorial !!

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mathmaniac

17 months ago, # |

Why isn't time limit for java 2x of C++ in codeforces, it is very painful to see same solutions running in C++ rather than in java, and this makes extremely difficult to pass solutions in java.

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BagishovMikail

17 months ago, # ^ |

It's your choice if every contestant, what language to use. If you choose C++, you get good performance, but loose, for instance, BigInteger. If you want write all problems using same language, it's definitely your problems.

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17 months ago, # |

← Rev. 2 →

"Let N be our initial set. Sadly we cannot just generate D(N) since its size might be of the order of 8·10^^8". How did the you compute the number 8.10^^8 ?

Also, similarly, how did you come with the number O(log 10^^18 .(|Da| + |Db|)) ?

Thanks.

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17 months ago, # ^ |

We cannot generate such a big set of numbers, so we are looking for a better way of computing the answer, and this way is described in the editorials.

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17 months ago, # ^ |

But how did we approximate the size of this big set of numbers?

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17 months ago, # ^ |

Use bruteforce to obtain those sets :)

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Single_Ready_To_Mingle

17 months ago, # ^ |

Complexity to generate those sets using bruteforce ?

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Lance_HAOH

17 months ago, # ^ |

O(k) where k is the number of integers ≤ 1e18 whose prime divisors are entirely contained in the set that you are generating.

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yhha

16 months ago, # ^ |

Can you tell me how to use bruteforce to calculate "D(N) which size might be of the order of 8 * 10^8 " please? Thanks.

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17 months ago, # |

Can someone elaborate on the implementation details of the problem E?

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17 months ago, # ^ |

Nevermind. I understood the implementation.

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17 months ago, # |

The first thought is to send the first elements to A and the rest to B. However, this is not enough; in this case the approximate size of D(A) might reach 7·106 which is way too much to pass.

How do we know this?

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Lance_HAOH

17 months ago, # ^ |

If you write a code to generate D(A) where A contains the first 8 prime numbers. You will see that its size will indeed be approximately 7e6. I have verified this on my computer.

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17 months ago, # |

What is the time complexity of the "count_le" function in x3n's code?

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JustLive

17 months ago, # ^ |

← Rev. 3 →

It's $\text{[math]}$ (or $\text{[math]}$ )

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17 months ago, # |

"The first thought is to send the first n / 2 elements to A and the rest to B. However, this is not enough".

I did exactly that and it passed in around 1100MS: http://codeforces.com/contest/912/submission/33953122

GreenGrape is that a typo, or did I get lucky?

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17 months ago, # ^ |

That’s not a typo :)
My solution uses two pointers approach and hence works in $\text{[math]}$

Yours iterates through the smaller set and binary searches the greater, and hence works in $\text{[math]}$

That means that a skew partition works fine in your case and fails in mine. As an opposite, the equallized partition slows down your solution and speeds up mine.

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17 months ago, # ^ |

Got it..thanks :)

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nikich340

17 months ago, # |

← Rev. 2 →

In problem C, isn't it easier just to add f(t) changes in map and then count ans after any change (choose max of them). And then output ans if f(t) == 0 after all changes or "-1" in another case. like this

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17 months ago, # |

← Rev. 2 →

For problem E: Can someone explain what's happening in the function "go" of the editorial solution? I understand that the primes are split between lp and rp, but I didn't get what the recursive function calls will do:

go(1, 0, lp, A);
go(1, 0, rp, B);

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17 months ago, # ^ |

It generates all numbers less than 10^18 which can be written as product of primes in the desired set.

For example primes = {2,3}, you generate all numbers less than 10^18 which can be written in the form 2^k1 * 3^k2

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Single_Ready_To_Mingle

17 months ago, # ^ |

What is the complexity to generate all such numbers in the "go" function ?

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I_love_Captain_America

17 months ago, # ^ |

complexity per prime = log(1e18)/log(2) * log(1e18)/log(3) * log(1e18)/log(next prime in the sequence) *.... * log(1e18)/log(current prime)

here log has base 10.

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I_love_Captain_America

17 months ago, # ^ |

But the interesting thing is, this is a very loose upper bound. So in practice it's not as bad as it looks.

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pk842

17 months ago, # |

thanks for a nice editorial :) specially D and E !

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nimom

17 months ago, # |

← Rev. 2 →

Can't E be solved using bitmask?

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de_vika

17 months ago, # ^ |

I thought of it too, but since the occurrence of each prime can be more than once, I decided against it..

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I_love_Captain_America

17 months ago, # |

I liked the thought behind E. It's all about optimizations!

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17 months ago, # |

some one interesting to problem B if you can choose "exactly" K numbers? obviously K=1 or 2 is like the ordinary problem

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17 months ago, # ^ |

I had the same thoughts — how would we solve problem B if the question was finding max XOR given we can pick "exactly k numbers"

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17 months ago, # |

In problem " B " , in testcase #48 , n was 2^58,so answer should be 2^59-1(xorring 2^58 and 2^58 -1) but answer was 2^58-1 (when k=2)!! how!! also why different answers with log2() and log2l()... is it about (rounding off) of double value

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17 months ago, # ^ |

← Rev. 3 →

2^58=1<<58=288230376151711744 testcase #48 is 288230376151711743 which is (2^58)-1 so...

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17 months ago, # ^ |

oops!!xD

thanks....

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17 months ago, # ^ |

but when i just used log2() instead of log2l() ,(int)(log2(n)+1) gave (59) and (int)(log2l(n)+1) gave 58!!why

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17 months ago, # ^ |

← Rev. 2 →

both to me i have 59.. i dont know why you have 58 for (int)(log2l(288230376151711744)+1) UPD you can submit this to any problem

#include <bits/stdc++.h>
using namespace std;
int main(){
	cout<<(int)(log2l(288230376151711744)+1)<<endl;
	cout<<(int)(log2(288230376151711744)+1)<<endl;
}

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17 months ago, # ^ |

pls check your code for 288230376151711743 = (2^58-1) giving 58 ,59!!!!!

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17 months ago, # ^ |

Really! its seem log2l work specially ,sor i dont kwow what log2l going on ,i usually use log2 instead maybe some "double" type problem ?

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mahim007

17 months ago, # |

For problem D, May I ask why this part || (f_a == f_b && a > b); is added in the compare function of solution 2? (f_a == f_b && a > b) and (f_a == f_b && a < b) both give AC and without these output do not match.I wonder why?

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17 months ago, # ^ |

A set cannot contain equal elements by its definition. Hence if f_a = f_b, the second point will not be added.

In order to fix it we should force our set to treat them as distinct ones. Since all points are distinct, this comparison works fine.

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mahim007

17 months ago, # ^ |

I still do not understand.

Since all points are distinct, then why do we need to force the set to treat them as distinct? Am I missing something?

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ruban

17 months ago, # |

In D we can use another method. Let see that optimal area is convex (or approach to convex). For this reason we should find rectangle area with square >=2*k. For find it let move vertical or gorizontal bounders (what's is better).

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17 months ago, # ^ |

← Rev. 2 →

Yeah, but most such solutions fail somewhere between test cases 81 and 95.

It’s cool that yours passed.

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AlwaysHustle

17 months ago, # |

in Tricky Alchemy problem,my algorithm was correct but in third test case, the output was correct but negative (-2147483648),i tried using long long,long double,int and it didn't work,any help?

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17 months ago, # ^ |

Please add a link to your solution below.

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AlwaysHustle

17 months ago, # ^ |

Here,http://codeforces.com/contest/912/submission/33944085

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andrewduffey

17 months ago, # ^ |

← Rev. 2 →

You should be using long long since 2147483648 is not a valid value for an int, it overflows resulting in the answer you are getting.

If you change the inputs to be long long and change the printf flag to %lld then it should pass.

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AlwaysHustle

17 months ago, # ^ |

Thank you so much you're right,i used long long but in scanf typed %ld,thanks a lot

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17 months ago, # |

"The first thought is to send the first elements to A and the rest to B. However, this is not enough; in this case the approximate size of D(A) might reach 7·106 which is way too much to pass. To speed it up we can, for example, comprise A of elements with even indexes (i.e p0, p2, ...) so that sizes of both A and B do not exceed 106 and the solution runs swiftly."

What's the difference? In both the case, one set will be of size floor(n/2), and the other will be of size n — floor(n/2). What am I missing?

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17 months ago, # ^ |

Applied a minor correction, thank you. Should be up soon :)

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17 months ago, # ^ |

done?

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17 months ago, # ^ |

pls help me out of weirdity of log2() and log2l() behaviour towards (2^58-1) in problem b:(

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kumarshobhit1998

17 months ago, # |

How do you get better? I'm always able to solve only one or two questions in div 2?

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porwalok

17 months ago, # |

Please help me out with Problem B \n Link to code- http://codeforces.com/contest/912/submission/33950052 \n I first convert number n to binary, then traverse the binary string checking if '0' is there -> convert it to '1' and decrement k. Getting WA from test case 9.

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_Mr_Fab

17 months ago, # |

Wonderful job with the editorial!!

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utsavgoel11

17 months ago, # |

In D, is it necessary that all fishes are placed horizontally only or verticaly only? The sample case doesn't cover the case of all 3 fishes under the square.

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HonestCad

17 months ago, # ^ |

The problem statement says you are not allowed to put more than one fish per cell.

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utsavgoel11

17 months ago, # ^ |

I know that, I'm saying that the 3 fishes can be placed in the shape of L in three different cells so that all 3 can be covered under the square

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utsavgoel11

17 months ago, # ^ |

← Rev. 2 →

"In the first example you can put the fishes in cells (2, 1), (2, 2), (2, 3)." Why so? why not (1,1), (2,1) , (2,2), so that output is 3?

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HonestCad

17 months ago, # ^ |

I believe it is because they are asking you to place fishes such that a random net has the highest expected value. In the given example, the expected value is 2 because there are 4 possible net positions and those net positions capture 8 fish --> Expected value = (total fish) / (net positions) = 8/4 = 2.

If you place the fish like you said at (1,1) (2,1) (2,2) it is true that the first net position (top left) will contain three fish. However, you will see that because of this the other net positions will contain 1, 2, and 1 fish. Then total fish = 3 + 1 + 2 + 1 = 7. Expected value = 7/4 which is less than 2 so this placement is not maximizing the expected value.

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JVirak

17 months ago, # |

So I've been stuck on E for a bit now (did this as a virtual contest and spent most of my time on it). I had a (I think) similar-ish idea. Instead of splitting the set in two I added one prime at a time, keeping track of the smallest k numbers one can make with the first m of the n primes. This (or at least my implementation of it) ended up being too slow. Is there a reason the DP approach is doomed to be too slow that I'm not seeing? Or is it likely just that my implementation sucks? Also (relatedly) I'm not seeing how to prove the O(|D(A)|+|D(B)|) part of the complexity? I suspect if someone would be nice enough to fill in this bit I'll see what I'm doing wrong on this problem.

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17 months ago, # ^ |

k might be as large as 8·10⁶.

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aj1

17 months ago, # ^ |

Hello GreenGrape, can you please explain how to estimate number of recursive calls that it takes to build sets D(A) and D(B)? And how do you come up with 10^6 limit? The way I estimate it is this: For the worst case we pick for A primes 2,5,11,17,23,31,41,47. Therefore total time is 18/log(2) * 18/log(5) ... 18/log(47)~10^9... Which will exceed time limit...

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pariz0

17 months ago, # |

Direction array:

int dx[4] = {-1 ,1 ,0 , 0} ;

int dy[4] = {0 ,0 -1 , 1} ; ==> gave wrong answer. but then

changing it to ,

int dx[4] = {-1 ,0 ,0 , 1} ;

int dy[4] = {0 ,-1 , 1 , 0} ; ==> the solution got accepted!

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pariz0

17 months ago, # |

In the problem, E.I'm stuck in bs implementation of @GreenGrape.

my question is,for smallest t we got count_le(t) == k.how can I prove that t is in the set of some combination of(D(A)*D(B))?

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yaksha

17 months ago, # |

GreenGrape , can you explain how do we get the numbers 8*10^8 (size of D(N) ) ... and also how do we get numbers 10^6 if we split N into A and B...

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bhishma

17 months ago, # ^ |

I think he found it out empirically . You can recursively compute the set A and B . It would be helpful if you try to think of it as a tree . At each height you can branch off into some power of the corresponding prime.

static void brute(int idx ,int len,  long curr, long collect[]) {
        if(idx == len) { 
            collect[ptr++] = curr;
            return;
        }
        brute(idx + 1, len , curr , collect);
        while(curr <= MAX / primes[idx]) {
            curr *= primes[idx];
            brute(idx + 1, len, curr, collect);
        }
    }

I'm currently stuck in a TLE while computing the the number of numbers less than a given number part . I tried Med0b1011's idea. If someone has solved E in java and faced the same issue please help me out . CODE

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17 months ago, # ^ |

← Rev. 3 →

I don't think it's possible to pass (logn)^2 in Java, at least without doing major changes to the generate function.

The solution in mashup takes 8190 seconds to run. The time limit is around 3.5 seconds, this looks like too much to optimize.
All Java solutions that passed run from 1.5 to 3.5 seconds, and they do it in logN. I don't think it would be possible to squeeze an extra log factor in.

The fastest solution in Java here runs in 1200MS : http://codeforces.com/contest/912/submission/33951400

Maybe you can use his generate function, and combine it with double binary search.

Note : I still find it surprisingly weird how can my CPP solution run in 1000MS without even attempting to optimize it while Java takes 8. I expected it to run in say 4-5 seconds at most.

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bhishma

17 months ago, # ^ |

← Rev. 2 →

I think that's due to using ArrayList , I wrote another solution using arrays but even that gave a TLE though it ran ~3.3s on my PC . Actually generate and sort took ~0.8s only. Can you try running this

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17 months ago, # ^ |

← Rev. 2 →

Takes 3.7 seconds, you are almost there. I think Java uses quick sort so that may slow it down (Changing long to Long to force merge sort takes 8) so try using your own sort function.

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bhishma

17 months ago, # ^ |

← Rev. 2 →

Thanks a lot , Finally AC

Used UWI's radix sort
Removed Java's binarySearch and implemented my own

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sumanyu

17 months ago, # |

Can someone please share a problem which is similar to problem D,where we need to find the first K values which maximizes or minimizes a given function. Thanks.. :)

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arsrivish2

17 months ago, # |

In this Code (rek): 33953146 what does cnt[at-1]+=0 and cnt[time-1]+=0 do? why to do += by zero

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17 months ago, # ^ |

← Rev. 2 →

That's a tricky (but not really beautiful in terms of code style) thing that lets us add a key (at-1 and time-1 respectively) to an STL C++ map. The "square brackets" operator of the map will create a key if it doesn't exist yet, so:

if we have no such key, it will simply add it and set its value to 0 so it won't affect answer
if we do have, the value won't get changed, so it won't affect the answer as well

Why do we need to add this key at all? It's because both at-1 and time-1 are the last points of times when we can kill a certain enemy before its health gets changed to be too high to let us kill it.

Though I personally think that it's ugly as I mentioned before, it's short and efficient :)

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arsrivish2

17 months ago, # ^ |

Thank you very much!!

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agamanan96

17 months ago, # |

← Rev. 2 →

Can someone explain C problem ? I couldn't understand the tutorial and why cnt[at]--?

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17 months ago, # ^ |

at is the point where enemy's health becomes too high to kill it, so we have to decrease count of the people we can kill there by 1.

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agamanan96

17 months ago, # ^ |

but intitially cnt[at]==0 ,right??

so it will get -1

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